How to decompose the following rational function into partial fractions?

Title: Partial Fraction Decomposition of $$ \frac{x^n}{(x-1)(x-3)(x-5)\cdots(x-(2n-1))} $$

Body:

I am working on decomposing the following rational function into partial fractions: $ \frac{x^n}{(x-1)(x-3)(x-5)\cdots(x-(2n-1))}. $

The denominator is a product of $ n $ odd linear factors $ (x-1), (x-3), \dots, (x-(2n-1)) $, and the numerator is $ x^n $. I want to express this as a sum of partial fractions in the form: $ \frac{x^n}{(x-1)(x-3)(x-5)\cdots(x-(2n-1))} = \sum_{k \in \{1, 3, 5, \ldots, 2n-1\}} \frac{A_k}{x-k}, $ where $ A_k $ are constants to be determined.

Using the standard method, I know that: $ A_k = \frac{k^n}{\prod_{\substack{j \neq k \\ j \in \{1, 3, 5, \ldots, 2n-1\}}} (k-j)}. $

Given the structure of the denominator with odd factors, I am curious:

Any help in deriving a more explicit or simplified relationship for $A_k $ would be greatly appreciated!

The following part of the problem is that:

Deduce the relation: $\sum_{k=0}^n (-1)^k (2k + 1)^n \binom{n}{k} = (-1)^n 2^n n!$, $\forall n \in \mathbb{N}$.

This is a hint in the right direction. Here are the partial fractions for some small values of n. Let $$P_n = \frac{x^n}{(x-1)(x-3)\cdots(x-(2n-1))}$$

$ P_1 = \frac{1}{x-1}+1 \\ P_2 = -\frac{1}{2 (x-1)}+\frac{9}{2 (x-3)}+1 \\ P_3 = \frac{1}{8 (x-1)}-\frac{27}{4 (x-3)}+\frac{125}{8 (x-5)}+1 \\ P_4 = -\frac{1}{48 (x-1)} +\frac{81}{16 (x-3)}-\frac{625}{16 (x-5)}+\frac{2401}{48 (x-7)}+1 \\ P_5 = \frac{1}{384 (x-1)} -\frac{81}{32 (x-3)} +\frac{3125}{64 (x-5)}-\frac{16807}{96 (x-7)}+\frac{19683}{128 (x-9)}+1 $

What patterns do you notice? I certainly notice alternating terms, perfect cubes in the numerator, powers of odd numbers, and some symmetry in the denominators of the terms, and a $1$ as the final term. Try to write a closed form that captures all of this.

The next step, of course, is proving it :)

Hope this helps.

$$P_n = \frac{(-2)^{-n} x^n}{\left(\frac{1}{2}-\frac{x}{2}\right)_n}$$ where $(a)_n$ is the Pochhammer symbol.

Hint:

$$\frac{x^5}{(x-1) (x-3) (x-5) (x-7) (x-9)}=1+\frac{1}{{2^4 4!}}\left( (-1)^0\binom{4}{0}\frac{1^5}{x-1} + (-1)^1\binom{4}{1}\frac{ 3^5}{x-3} + (-1)^2\binom{4}{2} \frac{5^5}{x-5} + (-1)^3\binom{4}{3}\frac{7^5}{x-7} + (-1)^4\binom{4}{4} \frac{9^5}{x-9} \right)$$

It suffices to apply the Lagrange interpolating polynomial to the function $f(x) =x^n$ at the nodes $\{1,3,...,(2n-1) \}$ and these $A_j$ you are seeking correspond to "these $y_j$ multiplying by the constant terms of the basis polynomials $l_j(x)$".